Problem: $\int(2x-5)^{10}dx\,=$ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{(2x-5)^{11}}{11}+C$ (Choice B) B $\dfrac{(2x-5)^{9}}{9}+C$ (Choice C) C $\dfrac{(2x-5)^{11}}{22}+C$ (Choice D) D $\dfrac{(2x-5)^{9}}{18}+C$
Explanation: If we let $ {u=2x-5}$, then ${du=2 \, dx}$ and $dx=\dfrac{du}{2}}$. Substituting gives us: $\begin{aligned}\int({2x-5})^{10}\,D {dx}\,&= \int u^{10}\,\cdot \dfrac{du}{2}}\,\\\\\\ &= \int u^{10}\cdot \dfrac12\, du\\\\\\ &=\dfrac12 \int u^{10}\,du\,\end{aligned}$ We recognize this antiderivative. $\begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx}&=\dfrac12 \int u^{10}\,du\,\\\\\\ &=\dfrac12\cdot \dfrac{u^{11}}{11}+C\\\\\\ &=\dfrac{u^{11}}{22}+C\end{aligned}$ We can now substitute back to find the antiderivative in terms of $x$. ∫ e x 1 + e 2 x d x = u 11 22 + C = ( 2 x − 5 ) 11 22 + C \begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx~}&=\dfrac{u^{11}}{22}+C\\\\\\\ &=\dfrac{(2x-5)^{11}}{22}+C\end{aligned} The answer: $\int(2x-5)^{10}dx\,=\dfrac{(2x-5)^{11}}{22}+C$